Verified answer

 y²-9....y+2

------ • ------ = y-3

y+2 ....y+3

(See below for work)

-

( y - 3 ) ( y + 3 ) ( y + 2 )

------------------------------------ = y - 3

( y + 2 ) ( y + 3 )

[(y² - 9)/(y + 2)] × [(y + 2)/(y + 3)]

= (y² - 9) / (y + 3)

= (y - 3)

(y² - 9/y + 2) × (y + 2/y + 3)

= y²(y + 2/y + 3) - 9(y + 2/y + 3)/y + 2(y + 2/y + 3)

= (y³ + 2y + 3y²) - (9 + 18/y² + 27/y) + (2y + 4/y + 6)

= y³ + 3y² + 2y + 2y - 9 + 6 - 27/y + 4/y - 18/y²

= y³ + 3y² + 4y - 2 - 23/y - 18/y²

y - 3 , y ╪ - 2 or - 3

(y² - 9) / (y + 2) * (y + 2) / (y + 3)

The (y + 2)'s cancel leaving:

(y² - 9) / (y + 3)

Then the difference of two squares can be factored leaving a pair of (y + 3)'s that cancel out:

(y + 3)(y - 3) / (y + 3)

y - 3

(y^2 -9) / (y +2) *(y+2) / (y+3)

(y^2 - 9)(y +2) / ((y+2)(y+3))

y+2 divided by y +2 = 1 where y does not equal -2

(y^2 -9) / (y +3)

(y+3)(y-3) / (y+3)

y+3 where y does not.equal -2 or 3